3.507 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=215 \[ \frac {a^{5/2} (163 A+200 B+304 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{64 d}+\frac {a^3 (299 A+392 B+432 C) \sin (c+d x)}{192 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (17 A+24 B+16 C) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{32 d}+\frac {a (5 A+8 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{24 d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d} \]
[Out]
1/64*a^(5/2)*(163*A+200*B+304*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+1/24*a*(5*A+8*B)*cos(d*x+
c)^2*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/4*A*cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+1/192*a^3*(299
*A+392*B+432*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/32*a^2*(17*A+24*B+16*C)*cos(d*x+c)*sin(d*x+c)*(a+a*sec(d
*x+c))^(1/2)/d
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Rubi [A] time = 0.70, antiderivative size = 215, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used =
{4086, 4017, 4015, 3774, 203} \[ \frac {a^3 (299 A+392 B+432 C) \sin (c+d x)}{192 d \sqrt {a \sec (c+d x)+a}}+\frac {a^{5/2} (163 A+200 B+304 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{64 d}+\frac {a^2 (17 A+24 B+16 C) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{32 d}+\frac {a (5 A+8 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{24 d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(a^(5/2)*(163*A + 200*B + 304*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(64*d) + (a^3*(299*A
+ 392*B + 432*C)*Sin[c + d*x])/(192*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(17*A + 24*B + 16*C)*Cos[c + d*x]*Sqrt
[a + a*Sec[c + d*x]]*Sin[c + d*x])/(32*d) + (a*(5*A + 8*B)*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d
*x])/(24*d) + (A*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(4*d)
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 3774
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Rule 4015
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[(A*b^2*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(a*f*n*Sqrt[a + b*Csc[e + f*x]]), x] +
Dist[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; Fr
eeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] &&
LtQ[n, 0]
Rule 4017
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2
- b^2, 0] && GtQ[m, 1/2] && LtQ[n, -1]
Rule 4086
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Rubi steps
\begin {align*} \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac {\int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (5 A+8 B)+\frac {1}{2} a (A+8 C) \sec (c+d x)\right ) \, dx}{4 a}\\ &=\frac {a (5 A+8 B) \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac {\int \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {3}{4} a^2 (17 A+24 B+16 C)+\frac {1}{4} a^2 (11 A+8 B+48 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {a^2 (17 A+24 B+16 C) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {a (5 A+8 B) \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac {\int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{8} a^3 (299 A+392 B+432 C)+\frac {1}{8} a^3 (95 A+104 B+240 C) \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac {a^3 (299 A+392 B+432 C) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (17 A+24 B+16 C) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {a (5 A+8 B) \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac {1}{128} \left (a^2 (163 A+200 B+304 C)\right ) \int \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (299 A+392 B+432 C) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (17 A+24 B+16 C) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {a (5 A+8 B) \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}-\frac {\left (a^3 (163 A+200 B+304 C)\right ) \operatorname {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}\\ &=\frac {a^{5/2} (163 A+200 B+304 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {a^3 (299 A+392 B+432 C) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (17 A+24 B+16 C) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {a (5 A+8 B) \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}\\ \end {align*}
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Mathematica [A] time = 2.33, size = 156, normalized size = 0.73 \[ \frac {a^2 \sqrt {\cos (c+d x)} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} \left (3 \sqrt {2} (163 A+200 B+304 C) \sin ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sin \left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} ((362 A+272 B+96 C) \cos (c+d x)+4 (23 A+8 B) \cos (2 (c+d x))+12 A \cos (3 (c+d x))+581 A+632 B+528 C)\right )}{384 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(a^2*Sqrt[Cos[c + d*x]]*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(3*Sqrt[2]*(163*A + 200*B + 304*C)*ArcSin[
Sqrt[2]*Sin[(c + d*x)/2]] + 2*Sqrt[Cos[c + d*x]]*(581*A + 632*B + 528*C + (362*A + 272*B + 96*C)*Cos[c + d*x]
+ 4*(23*A + 8*B)*Cos[2*(c + d*x)] + 12*A*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(384*d)
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fricas [A] time = 0.77, size = 444, normalized size = 2.07 \[ \left [\frac {3 \, {\left ({\left (163 \, A + 200 \, B + 304 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (163 \, A + 200 \, B + 304 \, C\right )} a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (48 \, A a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (23 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (163 \, A + 136 \, B + 48 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (163 \, A + 200 \, B + 176 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{384 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (163 \, A + 200 \, B + 304 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (163 \, A + 200 \, B + 304 \, C\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (48 \, A a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (23 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (163 \, A + 136 \, B + 48 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (163 \, A + 200 \, B + 176 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{192 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
[1/384*(3*((163*A + 200*B + 304*C)*a^2*cos(d*x + c) + (163*A + 200*B + 304*C)*a^2)*sqrt(-a)*log((2*a*cos(d*x +
c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(co
s(d*x + c) + 1)) + 2*(48*A*a^2*cos(d*x + c)^4 + 8*(23*A + 8*B)*a^2*cos(d*x + c)^3 + 2*(163*A + 136*B + 48*C)*a
^2*cos(d*x + c)^2 + 3*(163*A + 200*B + 176*C)*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*
x + c))/(d*cos(d*x + c) + d), -1/192*(3*((163*A + 200*B + 304*C)*a^2*cos(d*x + c) + (163*A + 200*B + 304*C)*a^
2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (48*A*a^2*cos
(d*x + c)^4 + 8*(23*A + 8*B)*a^2*cos(d*x + c)^3 + 2*(163*A + 136*B + 48*C)*a^2*cos(d*x + c)^2 + 3*(163*A + 200
*B + 176*C)*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)]
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giac [B] time = 4.05, size = 1542, normalized size = 7.17 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
-1/384*(3*(163*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 200*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 304*C*sqrt(-a)*a^2*sg
n(cos(d*x + c)))*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2
) + 3))) - 3*(163*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 200*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 304*C*sqrt(-a)*a^2
*sgn(cos(d*x + c)))*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqr
t(2) - 3))) + 4*sqrt(2)*(489*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^14*A*sqrt(-
a)*a^3*sgn(cos(d*x + c)) + 600*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^14*B*sqrt
(-a)*a^3*sgn(cos(d*x + c)) + 912*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^14*C*sq
rt(-a)*a^3*sgn(cos(d*x + c)) - 10269*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^12*
A*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 12600*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))
^12*B*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 19152*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 +
a))^12*C*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 69885*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)
^2 + a))^10*A*sqrt(-a)*a^5*sgn(cos(d*x + c)) + 103992*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1
/2*c)^2 + a))^10*B*sqrt(-a)*a^5*sgn(cos(d*x + c)) + 137424*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*
x + 1/2*c)^2 + a))^10*C*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 259233*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1
/2*d*x + 1/2*c)^2 + a))^8*A*sqrt(-a)*a^6*sgn(cos(d*x + c)) - 339864*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*t
an(1/2*d*x + 1/2*c)^2 + a))^8*B*sqrt(-a)*a^6*sgn(cos(d*x + c)) - 374544*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(
-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*C*sqrt(-a)*a^6*sgn(cos(d*x + c)) + 209979*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - s
qrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^7*sgn(cos(d*x + c)) + 262920*(sqrt(-a)*tan(1/2*d*x + 1/2*c)
- sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*sqrt(-a)*a^7*sgn(cos(d*x + c)) + 266928*(sqrt(-a)*tan(1/2*d*x + 1/
2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*sqrt(-a)*a^7*sgn(cos(d*x + c)) - 55511*(sqrt(-a)*tan(1/2*d*x +
1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*sqrt(-a)*a^8*sgn(cos(d*x + c)) - 73640*(sqrt(-a)*tan(1/2*d*
x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*sqrt(-a)*a^8*sgn(cos(d*x + c)) - 75888*(sqrt(-a)*tan(1/2
*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*C*sqrt(-a)*a^8*sgn(cos(d*x + c)) + 6687*(sqrt(-a)*tan(1
/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*sqrt(-a)*a^9*sgn(cos(d*x + c)) + 8808*(sqrt(-a)*tan
(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*sqrt(-a)*a^9*sgn(cos(d*x + c)) + 9456*(sqrt(-a)*t
an(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*sqrt(-a)*a^9*sgn(cos(d*x + c)) - 299*A*sqrt(-a)
*a^10*sgn(cos(d*x + c)) - 392*B*sqrt(-a)*a^10*sgn(cos(d*x + c)) - 432*C*sqrt(-a)*a^10*sgn(cos(d*x + c)))/((sqr
t(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(
-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^4)/d
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maple [B] time = 1.87, size = 1108, normalized size = 5.15 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^4*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
[Out]
1/3072/d*(489*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)^3*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/co
s(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+600*B*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(7/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+912*C*2^(1
/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+
c)))^(7/2)*sin(d*x+c)*cos(d*x+c)^3+1467*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*
x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+1800*B*sin(d*x+c)*cos(d*x+c)^
2*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos
(d*x+c)*2^(1/2))+2736*C*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2)
)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*sin(d*x+c)*cos(d*x+c)^2+1467*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)*arctanh(1/
2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+180
0*B*sin(d*x+c)*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+2736*C*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d
*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*sin(d*x+c)*cos(d*x+c)+489*A*2^(1/2)*arctanh(1/2
*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*sin(
d*x+c)+600*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin
(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)+912*C*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+
c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*sin(d*x+c)-768*A*cos(d*x+c)^8-2176*A*cos(d*x+c)^7-
1024*B*cos(d*x+c)^7-2272*A*cos(d*x+c)^6-3328*B*cos(d*x+c)^6-1536*C*cos(d*x+c)^6-2608*A*cos(d*x+c)^5-5248*B*cos
(d*x+c)^5-6912*C*cos(d*x+c)^5+7824*A*cos(d*x+c)^4+9600*B*cos(d*x+c)^4+8448*C*cos(d*x+c)^4)*(a*(1+cos(d*x+c))/c
os(d*x+c))^(1/2)/cos(d*x+c)^3/sin(d*x+c)*a^2
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(c + d*x)^4*(a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
[Out]
int(cos(c + d*x)^4*(a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**4*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
[Out]
Timed out
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